1) The point of this question is to start out from the standard equation: ax²+bx+c=0 to get the quadratic equation:
[tex]x=\frac{b\pm\sqrt{b^2-4ac}}{2a}[/tex]2) So, let's do it step by step so that we can get what's going to be Jeff's next step:
[tex]\begin{gathered} I)ax^2+bx+c=0 \\ II)ax^2+bx+c-c=-c \\ III)\frac{ax^2}{a}+\frac{bx}{a}=-\frac{c}{a} \\ IV)x^2+\frac{bx}{a}+\frac{c}{a}=0 \\ \\ \end{gathered}[/tex]Note that since the leading coefficient is a, we can divide both sides by
a. And the next step is to complete the square so that we can rewrite that as a perfect square trinomial.
3)
[tex]\begin{gathered} x^2+\frac{bx}{a}=-\frac{c}{a}\Rightarrow(\frac{b}{2a})^2 \\ \\ V)x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2} \\ \\ \end{gathered}[/tex]Notice that to complete the square, we had to square the coefficient b/2a
to both sides and add to both sides of the equation.